How many grams of sodium benzoate, C H CO Na, have to be added to 1.50 L of a 0.0200 M solution of benzoic acid, C H CO H, to make a buffer with a pH = 4.00, assuming no volume change?

Question

Skyla Hendricks

Chemistry

1 April, 19:56

How many grams of sodium benzoate, C H CO Na, have to be added to 1.50 L of a 0.0200 M solution of benzoic acid, C H CO H, to make a buffer with a pH = 4.00, assuming no volume change?

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Sunshine · Answer 1

Answer: 2.8275grams

Explanation: A buffer is made btw a weak acid and it salt. In a solution made by dissolving a weak acid in solution, equilibrium is set up btw ionised and unionised ion. For Benzoic acid

C6H5COOH ... > C6H5COO - + H+

Ka = [C6H5COO-] [H+] / [C6H5COOH] ... (1)

using Ka = 6.5 * 10^-5, [C6H5COOH] = 0.02M. PH = - log[H+] ... > [H+] = 10^-4M.

Putting the values in (1)

[C6H5COO-] = 6.5 * 10^-5 * 0.02 / 10^-4

[C6H5COO-] = 0.013M = Molarity of sodium benzoate

Mole (C6H5COONa) = 0.013 * Volume = 0.013mol/litre * 1.5 litre

Mole (C6H5COONa) = 0.0195mol

Mass (C6H5COONa) = 0.0195 * Molar mass

Mass (C6H5COONa) = 2.8275g